What is the extraneous solution to these equations? $\dfrac{x^2 - 3x}{x + 8} = \dfrac{-8x - 6}{x + 8}$
Answer: Multiply both sides by $x + 8$ $ \dfrac{x^2 - 3x}{x + 8} (x + 8) = \dfrac{-8x - 6}{x + 8} (x + 8)$ $ x^2 - 3x = -8x - 6$ Subtract $-8x - 6$ from both sides: $ x^2 - 3x - (-8x - 6) = -8x - 6 - (-8x - 6)$ $ x^2 - 3x + 8x + 6 = 0$ $ x^2 + 5x + 6 = 0$ Factor the expression: $ (x + 2)(x + 3) = 0$ Therefore $x = -2$ or $x = -3$ The original expression is defined at $x = -2$ and $x = -3$, so there are no extraneous solutions.